return to homepage The Bohr Magneton by Miles Mathis
Why is the Bohr Magneton not equal to the measured magnetic moment of the electron? In experiment, we find that the values are off by .1%. QED has no simple answer for this. I do. QED proposes to explain the error by once again pouring Dirac’s virtual sea on the problem and once again waving the magic wand. The electron is said to be interacting with virtual photons, giving it a precession and thereby a The Bohr magneton was first proposed by Procopiu in 1913. It is not a particle, but rather an expression of the magnetic field created by the individual electron. We have a simple equation for it: μ_{B} = eh/2π2m_{e}Unfortunately, as I said, this gives us a number that fails by about .1%. To see why, we must study this equation more closely. We can do this by looking at angular momentum, and the easiest way to do that is by returning to Bohr’s simple math, which I first critiqued in another paper. L = nh/2π = rmv We let n = 1, since we are studying the first electron in the hydrogen atom. So, μ_{B} = eL/2m = _{e}erv/2
We find that this equation yields the wrong number. Why? Because the math is wrong. As I showed in my paper called _{B} = erv/2
v = 2μ v = 2.19 x 10 Mirroring current assumptions, I used the Bohr radius for r. That is well under c. Why? Why doesn’t the electron maximize its orbital speed? I will be told that it is because v is not the orbital speed, ω is. Well, v = rω. So, ω = 4.14 x 10^{16}/s
Is that over c? Nobody knows, because nobody understands angular speed. You can’t just multiply or divide by a radius to make a linear velocity into an angular velocity. That doesn’t make any sense, mathematically or mechanically. Look at the equations for momentum and angular momentum closely: p = mv L = rmv If the radius is greater than one, the effective angular velocity will be more than the linear velocity. If the radius is less than one, the effective angular velocity will be less than the linear velocity. That is a flagrant example of illogical scaling. The history of physics fudges over this problem by creating a moment of inertia, but the moment of inertia is a ghost. It is the attempt to hide the fact that v = rω is wrong. You would not have a moment of inertia without v = rω. Where does that equation come from? It comes from 2πr/t. If v = 2πr/t, and ω = 2π/t, then v must equal rω. But, as I have shown, v ≠ 2πr/t. In the historical derivations, v is defined as the tangential velocity. But 2πr/t is not the tangential velocity; it is the orbital velocity. The orbital velocity curves and the tangential velocity does not. The tangential velocity is a straight line vector with its tail on the curve, but it does not follow the curve. If v = 2πr/t, then v is None of the angular momentum equations in books make any sense, so I developed my own equation to do this, going back to Newton to find the method. You can see my derivation in my paper on a = v ω = √[2r√v ^{2} + r^{2}) - 2r^{2}]
This equation is logical, because using it we find that the angular velocity is always less than the tangential velocity. We don’t have any misdirection with moments of inertia, and we don’t have the illogic of having the variables change in different ways for different values of r. We have a logical progression, since as we get larger, the angular velocity approaches the tangential velocity. Obviously, this is because it loses it curvature as it increases, becoming more like the straight-line vector. By the same token, at small scales, the angular velocity gets very small compared to the tangential velocity, and this is because the curvature is so great. Using my new equation for ω, L = mω = h/2π μ_{B} = eω/2
Now we just solve ω = 2(9.283 x 10^{-24})/1.602 x 10^{-19}ω
r = √[ω_{e} = 1.16 x 10^{-4}m/s^{4}/(4v^{2} - 4ω^{2})]
If we use c for v, we find, r Of course, I have redefined the variable ω here. It is no longer measured in radians. Strictly, it stands for the orbital velocity measured in meters, not the angular velocity measured in radians. My new equations simply separate it from the tangential velocity, since, as you have seen, we need both. We don't need to be able to measure circular motion in radians, but we do need to measure it both as a tangential velocity and as an orbital velocity. Therefore, I have jettisoned the old angular velocity in radians as a useless concept.
I will be told that the orbit must have a momentum of its own, angular or otherwise. Yes, there must be orbital energy, but we need not calculate an angular momentum. The only thing in the orbit is the electron, so there is no mass inside the sphere. The only thing inside the orbit is the nucleus, and it is not moving relative to the orbit, so it creates no angular momentum. This being true, we only have to look at mass and momentum at the tangent. If we do that, we don’t have to be concerned with angular anything. The electron transmits energy from the tangent, by emission, and this emission is emitted during a very small interval of the orbit. The emission leaves the electron in a straight-line vector, so we don’t have angular momentum involved. We can use the tangential velocity directly, and compute the momentum linearly, with p = mc. But this momentum does not contribute to the magnetic field. This momentum contributes to the linear energy of the emission, not its spin energy. Magnetism is not caused by linear energy, it is caused by spin energy. So we don’t have to be concerned with the orbital energy. Now, the electron does interact with the field outside the orbit, but this is not a virtual field. It is the emission field of other quanta. The vacuum is awash with emission, and this emission acts as a friction on the orbit. But this doesn’t cause a precession or a p = mc = 2.7 x 10 The outer spin has an opposite momentum of about 10 But why is the experimental number for the Bohr magneton .1% wrong? Is it just that Bohr's numbers were different than current numbers? Is it a problem of the virtual field, explained by the Finally, let’s check that value for the electron radius. Actually, what I found above is the radius of the outer spin. The electron in orbit has both an axial spin and an x-spin. Therefore the radius of the electron proper is: r But the x-spin radius, 2.244 x 10 r = 5.45 x 10 Which is very close. We can use my number to re-estimate the radius for the proton, assuming it has the same density as the electron. r We can fine tune that as well. Since we are finding radius here, not mass, we can use my spin equations from my paper unifying the electron and proton. In a nutshell, we find we need the transform 1822, not 1836. The number 1836 is a sort of mass transform, which means it must be a unified field transform. Mass is always a unified field number. But radius is not a unified field number. For that reason, we can just use the Dalton, 1822, which transforms size but not mass. In that case we get the number 4.09 x 10 for the proton radius. That is very interesting because it is the square root of the proton mass. See my new paper on proton mass for more on this.
^{-14}mOf course this means the Bohr radius is wrong as well. Bohr’s math is completely compromised by now, so everything has to be redone. The problem with angular velocity has infected all the math, and nothing will stand. Let’s correct the Bohr equation: mv First of all, a ≠ v a = ω Which gives us, mω
Using Bohr’s method, we find L = mω = h/2π ω = h/2πm hω/4π = ke ^{2}/hr
h/2πm = 4πke r = 8π This gives us the Think of the orbit like a big spinning particle, of radius r. That big particle has an angular momentum. The electron also has an angular momentum. Bohr has conflated the two. His equations are a mixing of both values. So he has made two big errors. One, he has used the wrong equation for angular momentum, based on a mistaking of tangential and orbital velocity. Two, he has a false equality. If our first equation (mω We will also see in subsequent papers that Schrodinger’s equations do not solve this problem. Schrodinger has Bohr’s principal quantum number and also a separate angular momentum quantum number, but he does not assign these physically. Because we don’t get the mechanics, the math is unclear. And Schrodinger’s equations retain the errors of Bohr in going from linear to angular velocity. That is, Schrodinger still uses a false angular momentum equation. L = rmv was not corrected by Schrodinger, and it has never been corrected since. Can we still find a Bohr radius? Let us assume that the first equation is right, after correcting the momentum equation. mω But we have two unknowns, r and ω, and only one equation. We can’t solve without another equation, and Bohr’s momentum equations are false. Let us first try using c for v. We will assume the tangential velocity of the electron is maximized. So we simply return to the equation ω = √[2r√v ^{2} + r^{2}) - 2r^{2}]
mω r√c We can simplify that by noticing that the left side will be dominated by c, allowing us to omit values of r. cr = ke r = √ke That’s way too large, so we know something is still wrong. Let’s try using Bohr’s radius to find the angular velocity. ω = √2ke^{2}/mr] = 3.09 x 10^{6}m/s
Interesting that that is almost what we got for the r v = ω So that can’t be right either. From these calculations, it would appear that the Bohr radius is either greater than or equal to about a millimeter, or we are using the wrong values for the electron, or the equation is still wrong. It turns out that the equation is still wrong. The problem is simple: the constant k doesn’t apply at the quantum level. Coulomb’s equation is for use at the macro-level, and the constant is a scaling constant. Just as I showed with G in another paper, k takes us from one level of size to another, so that we can compare fields that have different mediating particles or accelerations. Coulomb was working with little pith balls, not electrons, and his balls were nine orders of magnitude larger than the orbital radius of the electron, as you will see. I showed that G is a scaling constant that takes us from the size of emitted photons to the size of the atom. Yes, the Since in Bohr’s equation the field is the actual field the electron is moving in, we don’t need a transform or a scaling constant. The electron is already moving at the proper scale. In the little illustration in the book, we see the electron circling the nucleus, and the electron and the orbital radius are in the same field. We have to do very little scaling (between the charge and the field it is in) in order to draw the picture, and this is not beside the point. The proton is actually repulsing the electron down that very radius. And so we get this very simple equation: r = √( That is the corrected Bohr radius. The value of Coulomb’s constant is 9 x 10 What all this means is that Coulomb's constant is not a constant. It is taking us from one size to another, so it cannot be applied across a range of sizes. This was to be expected, since I have shown that Coulomb's equation, like Newton's equation, is a unified field equation that includes both the E/M field and the gravitational field. Furthermore, I have shown that the two fields are not the same relative size to eachother, as we scale the equations up and down. So the connection between Coulomb’s constant and the Bohr radius is not a coincidence. Although the current numbers are wrong, it is no coincidence that the current Bohr diameter is thought to be about 1/k meters. It is the mistake in the Coulomb equation that led directly to the mistake in the Bohr radius, and they are connected both mathematically and historically. It is also not a coincidence in my new math, since if you multiply the old Bohr radius and the real diameter of Coulomb’s ball by 177, you get my new Bohr radius and 1 meter. For more on this, see my paper on Coulomb's equation.
I have already shown that a misreading of the scattering equations means we have the atomic size about 100 times too small, so my new equation also fits that prediction and correction very well. The Bohr radius is 177 times larger than we thought, and you can now see all the math and logic behind the correction. If this paper was useful to you in any way, please consider donating a dollar (or more) to the SAVE THE ARTISTS FOUNDATION. This will allow me to continue writing these "unpublishable" things. Don't be confused by paying Melisa Smith--that is just one of my many |